Divide the following complex numbers. $ \dfrac{-3-3i}{3+3i}$
Explanation: We can divide complex numbers by multiplying both numerator and denominator by the denominator's complex conjugate , which is ${3-3i}$ $ \dfrac{-3-3i}{3+3i} = \dfrac{-3-3i}{3+3i} \cdot \dfrac{{3-3i}}{{3-3i}} $ We can simplify the denominator using the fact $(a + b) \cdot (a - b) = a^2 - b^2$ $ \dfrac{(-3-3i) \cdot (3-3i)} {(3+3i) \cdot (3-3i)} = \dfrac{(-3-3i) \cdot (3-3i)} {3^2 - (3i)^2} $ Evaluate the squares in the denominator and subtract them. $ \dfrac{(-3-3i) \cdot (3-3i)} {(3)^2 - (3i)^2} = $ $ \dfrac{(-3-3i) \cdot (3-3i)} {9 + 9} = $ $ \dfrac{(-3-3i) \cdot (3-3i)} {18} $ Note that the denominator now doesn't contain any imaginary unit multiples, so it is a real number, simplifying the problem to complex number multiplication. Now, we can multiply out the two factors in the numerator. $ \dfrac{({-3-3i}) \cdot ({3-3i})} {18} = $ $ \dfrac{{-3} \cdot {3} + {-3} \cdot {3 i} + {-3} \cdot {-3 i} + {-3} \cdot {-3 i^2}} {18} $ Evaluate each product of two numbers. $ \dfrac{-9 - 9i + 9i + 9 i^2} {18} $ Finally, simplify the fraction. $ \dfrac{-9 - 9i + 9i - 9} {18} = \dfrac{-18 + 0i} {18} = -1 $